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Answer: 323.61 g of 
will be produced
Explanation:
The given balanced chemical reaction is :
According to stoichiometry :
2 moles of 
require 1 mole of 
Thus 3.00 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product.
As 2 moles of give = 2 moles of
Thus 3.00 moles of give =
of
Mass of 
Thus 323.61 g of will be produced from the given moles of both reactants.
The ways in which ammonia can be identified is mentioned in below pointers.
<h3>What is Ammonia ?</h3>
Ammonia is a compound of nitrogen and hydrogen with the formula NH3.
A stable binary hydride, and the simplest pnictogen hydride.
Ammonia is a colourless gas with a distinct pungent smell.
Biologically, it is a common nitrogenous waste,
Three ways in which ammonia gas can be identified is:
- It has a sharp characteristic odor.
- When a glass rod dipped in HCl is brought in contact with the gas white color fumes of ammonium chloride are formed.
- It turns moist red litmus blue, moist turmeric paper brown, and phenolphthalein solution pink.
To know more about Ammonia
brainly.com/question/17198636
#SPJ4
It reflects sunlight. It's a physical barrier (in comparison to some sunscreens which are chemical blocks). Sunlight is mostly white light and zinc oxide reflects white light
The dissociation constant of the base : 7.4 x 10⁻⁴
<h3>Further explanation</h3>
Butylamine, C4H9NH2 Is A Weak Base
Kb is the dissociation constant of the base.
LOH (aq) ---> L⁺ (aq) + OH⁻ (aq)
![\rm Kb=\dfrac{[L][OH^-]}{[LOH]}](https://tex.z-dn.net/?f=%5Crm%20Kb%3D%5Cdfrac%7B%5BL%5D%5BOH%5E-%5D%7D%7B%5BLOH%5D%7D)
[OH⁻] for weak base can be formulated :
![\tt [OH^-]=\sqrt{Kb.M}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7BKb.M%7D)
pH of solution : 12
pH+pOH=14, so pOH :
14-12 = 2, then :
![\tt [OH^-]=10^{-pOH}\\\\(OH^-]=10^{-2}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D10%5E%7B-pOH%7D%5C%5C%5C%5C%28OH%5E-%5D%3D10%5E%7B-2%7D)
the the dissociation constant (Kb) =
Or you can use from ICE method :
C4H9NH2(aq) + H2O(l) ⇌ C4H9NH3+(aq) + OH-(aq)
0.15
x x x
0.15-x x x
![\tt Kb=\dfrac{x^2}{0.15-x}\rightarrow x=[OH^-]\\\\Kb=\dfrac{10^{-4}}{0.15-10^{-2}}=7.14\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20Kb%3D%5Cdfrac%7Bx%5E2%7D%7B0.15-x%7D%5Crightarrow%20x%3D%5BOH%5E-%5D%5C%5C%5C%5CKb%3D%5Cdfrac%7B10%5E%7B-4%7D%7D%7B0.15-10%5E%7B-2%7D%7D%3D7.14%5Ctimes%2010%5E%7B-4%7D)
